Calculate the concentration (in mol/l) of 33% by weight (33 g naoh per 100 g of solution) naoh solution. (the density of the 33%
1 answer:
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Answer:
After the transfer the pressure inside the 20 L vessel is 0.6 atm.
Explanation:
Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:
Therefore, for this problem the step by step explanation is:
Clearing P2 and replacing
Explanation:
The mass of a pot is 300g and contains 90% aluminum. Find the number of moles of aluminum in the pot. P.A. (Al = 27)
The mass of aluminum present in the pot is:
Hence, in the given pot 270g Al is present.
The gram atomic mass of Al -27 g/mol
Given the mass of Al is 270 g
Substitute these values in the above formula:
Answer is 10.0 mol of Al is present.
Answer:
a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) 0.0035 mole
c) 0.166 M
Explanation:
Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.
The equation of the reaction is expressed as:
1 mole 3 mole
The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows
10 ml 17.50 ml
(x) M 0.200 M
Molarity =
= 0.0035 mole
c) What was the molar concentration of phosphoric acid in the original stock solution?
By stoichiometry, converting moles of NaOH to H₃PO₄; we have
=
= 0.00166 mole of H₃PO₄
Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:
Molar Concentration =
Molar Concentration =
Molar Concentration = 0.166 M
∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M